Key Concepts and Formulas

• Modular Arithmetic: $a \equiv b \pmod{n}$ means $a$ and $b$ have the same remainder when divided by $n$.
• Euler's Totient Theorem: If $\text{gcd}(a, n) = 1$, then $a^{\phi(n)} \equiv 1 \pmod{n}$.
• Euler's Totient Function: For a prime power $p^k$, $\phi(p^k) = p^k - p^{k-1}$.

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Step-by-Step Solution

Step 1: Express the given information mathematically. We are given that the $50^{\text {th }}$ root of $x$ is 12, which can be written as $x^{1/50} = 12$. To find $x$, we raise both sides to the power of 50: $x = 12^{50}$ Similarly, for $y$, we have $y^{1/50} = 18$, which implies: $y = 18^{50}$ We need to find the remainder of $(x+y)$ when divided by 25, which is $(12^{50} + 18^{50}) \pmod{25}$.

Step 2: Calculate Euler's totient function for the modulus. The modulus is $n=25$. We need to find $\phi(25)$. Since $25 = 5^2$ (a prime power), we use the formula $\phi(p^k) = p^k - p^{k-1}$: $\phi(25) = \phi(5^2) = 5^2 - 5^{2-1} = 25 - 5 = 20$ Thus, $\phi(25) = 20$. This means for any integer $a$ coprime to 25, $a^{20} \equiv 1 \pmod{25}$.

Step 3: Verify coprimality of the bases with the modulus. We must check if the bases, 12 and 18, are coprime to the modulus, 25.

• For 12 and 25: The prime factors of 12 are 2 and 3. The prime factor of 25 is 5. They share no common factors, so $\text{gcd}(12, 25) = 1$.
• For 18 and 25: The prime factors of 18 are 2 and 3. The prime factor of 25 is 5. They share no common factors, so $\text{gcd}(18, 25) = 1$. Since both bases are coprime to 25, we can apply Euler's Totient Theorem.

Step 4: Simplify $12^{50} \pmod{25}$. We need to find $12^{50} \pmod{25}$. Using Euler's Totient Theorem, we know $12^{20} \equiv 1 \pmod{25}$. We reduce the exponent 50 modulo $\phi(25)=20$: $50 = 2 \times 20 + 10$ So, $12^{50} = 12^{2 \times 20 + 10} = (12^{20})^2 \times 12^{10}$ Applying the congruence: $12^{50} \equiv (1)^2 \times 12^{10} \pmod{25}$ $12^{50} \equiv 12^{10} \pmod{25}$ Now we calculate $12^{10} \pmod{25}$ by computing powers step-by-step:

• $12^1 \equiv 12 \pmod{25}$
• $12^2 = 144 \equiv 19 \pmod{25}$ (since $144 = 5 \times 25 + 19$)
• $12^4 \equiv (12^2)^2 \equiv 19^2 = 361 \pmod{25}$ (since $361 = 14 \times 25 + 11$) $\equiv 11 \pmod{25}$
• $12^5 \equiv 12^4 \times 12^1 \equiv 11 \times 12 = 132 \pmod{25}$ (since $132 = 5 \times 25 + 7$) $\equiv 7 \pmod{25}$
• $12^{10} \equiv (12^5)^2 \equiv 7^2 = 49 \pmod{25}$ (since $49 = 1 \times 25 + 24$) $\equiv 24 \pmod{25}$ Since $24 \equiv -1 \pmod{25}$, we have: $12^{50} \equiv -1 \pmod{25}$

Step 5: Simplify $18^{50} \pmod{25}$. We follow the same procedure for $18^{50} \pmod{25}$. The exponent reduction is the same: $50 = 2 \times 20 + 10$. $18^{50} = 18^{2 \times 20 + 10} = (18^{20})^2 \times 18^{10}$ Using Euler's Totient Theorem ($18^{20} \equiv 1 \pmod{25}$): $18^{50} \equiv (1)^2 \times 18^{10} \pmod{25}$ $18^{50} \equiv 18^{10} \pmod{25}$ Now we calculate $18^{10} \pmod{25}$:

• It's often easier to use negative remainders: $18 \equiv 18 - 25 \equiv -7 \pmod{25}$.
• $18^2 \equiv (-7)^2 = 49 \pmod{25}$.
• As calculated before, $49 \equiv 24 \equiv -1 \pmod{25}$. Now we can write $18^{10}$ as $(18^2)^5$: $18^{10} \equiv (18^2)^5 \equiv (-1)^5 \pmod{25}$ $(-1)^5 = -1$ Therefore, $18^{50} \equiv -1 \pmod{25}$

Step 6: Combine the results and find the final remainder. We need to find the remainder of $(x+y)$ when divided by 25. $(x+y) \equiv 12^{50} + 18^{50} \pmod{25}$ Substitute the results from Steps 4 and 5: $(x+y) \equiv (-1) + (-1) \pmod{25}$ $(x+y) \equiv -2 \pmod{25}$ To express this as a positive remainder (between 0 and 24), we add the modulus: $-2 + 25 = 23$ Thus, the remainder obtained on dividing $(x+y)$ by 25 is 23.

The final answer is $\boxed{1}$.